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Hardy-Weinberg Equilibrium Problem Set - Evolutionary Biology, Study notes of Evolutionary biology

A problem set focused on the hardy-weinberg equilibrium (hwe) for various genetic traits. It includes calculations of genotype and allele frequencies, as well as predictions based on the hwe law. The problem set covers topics such as thalassemia, pku, and blood types.

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2021/2022

Uploaded on 09/27/2022

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PROBLEM SET 1 - EVOLUTIONARY BIOLOGY - FALL 2016 KEY
Hardy-Castle-Weinberg (20 pts total)
1. A population of infants from Musoma, Tunisia, was scored for the β-hemoglobin locus genotype (determined
by a single autosomal locus with two alleles, A and S) as follows:
Genotype AA AS SS Total
Number 189 89 9 287
Characterize this population by its genotype frequencies. Characterize the gene pool by the allele frequencies
for A and S. Using Hardy-Weinberg Law, predict the genotype frequencies based on the allele frequencies. (3
pts)
Observed Genotype Frequencies:
Freq(AA) = 189/287 = 0.66
Freq(AS) = 89/287 = 0.31
Freq(SS) = 9/287 = 0.03
Allele Frequencies:
Freq(A) = p = 189/287 + ½(89/287) = 0.81
Freq(S) = q = 9/287 + ½(89/287) = 0.19
(Or use p + q = 1 to solve for q)
Expected HWE genotype frequencies
p2 = (0.81)2 = 0.66
2pq = 2(0.81)(0.19) = 0.3
q2 = (0.19)2 = 0.03
2. Among people of southern Italian and Sicilian ancestry living in Rochester, New York (N = 10,000), about one
birth in 2,500 has thalassemia major (a type of anemia) and about one birth in 25 has a milder anemia known
as thalassemia minor. Are these data compatible with a single locus hypothesis as a basis for the heredity of
these anemic conditions? Show why or why not. (3 pts)
To answer this question, construct a model with all the elements of the basic Hardy-Weinberg model and see
if the data you have been given fit. Assume two alleles and one locus and arbitrarily name the alleles A and a.
If you assume that thalassemia major is a homozygous genotype (AA) and that thalassemia minor is a
heterozygous genotype (Aa) then you have the genotype frequencies from the question to work with:
AA = 1/2500 = 0.0004; Aa = 1/25 = 0.04; and aa= 1-0.04-0.0004 = 0.9596
Solve for the allele frequencies:
pA= 0.0004 +1/2(0.04) = 0.0204
qa= 0.9596 + 1/2(0.04) = 0.9796
Given these values for p and q, the expected genotype frequencies under H-W are p2 = 0.0004; 2pq = 0.04;
q2 = 0.9596. These frequencies match the given values in the population. Therefore, this trait fits the H-W
two-allele, one-locus model.
3. Two populations are examined for the same gene locus with the following results:
pf3

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PROBLEM SET 1 - EVOLUTIONARY BIOLOGY - FALL 2016 KEY

Hardy-Castle-Weinberg (20 pts total)

  1. A population of infants from Musoma, Tunisia, was scored for theby a single autosomal locus with two alleles, A and S ) as follows: β -hemoglobin locus genotype (determined

Genotype AA AS SS Total Number 189 89 9 287 Characterize this population by its genotype frequencies. Characterize the gene pool by the allele frequenciesfor A and S. Using Hardy-Weinberg Law, predict the genotype frequencies based on the allele frequencies. ( pts) Observed Genotype Frequencies: Freq(AA) = 189/287 = 0.66Freq(AS) = 89/287 = 0. Freq(SS) = 9/287 = 0. Allele Frequencies: Freq(A) =Freq(S) = pq = 189/287 + ½(89/287) = 0.81= 9/287 + ½(89/287) = 0. (Or use p + q = 1 to solve for q ) Expected HWE genotype frequencies p 22 pq^ = (0.81) = 2(0.81)(0.19) = 0.3 2 = 0. q^2 = (0.19) 2 = 0.

  1. Among people of southern Italian and Sicilian ancestry living in Rochester, New York (N = 10,000), about onebirth in 2,500 has thalassemia major (a type of anemia) and about one birth in 25 has a milder anemia known as thalassemia minor. Are these data compatible with a single locus hypothesis as a basis for the heredity ofthese anemic conditions? Show why or why not. (3 pts) To answer this question, construct a model with all the elements of the basic Hardy-Weinberg model and see if the data you have been given fit. Assume two alleles and one locus and arbitrarily name the allelesIf you assume that thalassemia major is a homozygous genotype (AA) and that thalassemia minor is a A and a. heterozygous genotype (Aa) then you have the genotype frequencies from the question to work with: AA = 1/2500 = 0.0004; Aa = 1/25 = 0.04; and aa = 1-0.04-0.0004 = 0. Solve for the allele frequencies: p q (^) A = 0.0004 +1/2(0.04) = 0. a = 0.9596 + 1/2(0.04) = 0. Given these values for p and q , the expected genotype frequencies under H-W are p 2 = 0.0004; 2pq = 0.04; q two-allele, one-locus model. 2 = 0.9596. These frequencies match the given values in the population. Therefore, this trait fits the H-W
  2. Two populations are examined for the same gene locus with the following results:

Population 1:Population 2: AAAA 162;18; Aa Aa 84; 36; aa aa 98 2 Now suppose these two populations are combined to form a new population: Population 3: AA 180; Aa 120; aa 100 a. Can the differences in genotype frequencies in populations 1 and 2 be explained by differences in the mating systems with respect to this gene locus? (2 pts) Pop 1 Genotype frequencies AA = 162/200 = 0.81; Aa = 36/200 = 0.18; aa = 2/200 = 0.01Allele frequencies p Expected H-W frequencies A^ = 0.81 +1/2(0.18) = 0.9; p 2 = (0.9) 2 = 0.81; 2^ qpq^ a = 0.01 + 1/2(0.18) = 0.1 = 2(0.9)(0.1) = 0.18; q 2 = (0.1)^2 = 0. This population is in HW equilibrium Pop 2 Genotype frequencies AA = 18/200 = 0.09; Aa = 84/200 = 0.42; aa = 98/200 = 0. Allele frequenciesExpected H-W frequencies p (^) A = 0.09 +1/2(0.42) = 0.03; p 2 = (0.3) 2 = 0.09; 2 qpq (^) a = 0.49 + 1/2(0.42) = 0.7 = 2(0.3)(0.7) = 0.42; q 2 = (0.7) (^2) = 0. This population is in HW equilibrium Since both populations are in HW expected proportions they do not violate the assumption of randommating. There is no evidence that differences in the mating system are responsible for the differences in genotype frequencies. b. Is population 3 in H-W equilibrium? If not, indicate whether heterozygotes are in excess or indeficiency. (2 pts) Pop 3 Genotype frequencies AA = 180/400 = 0.45; Aa = 120/400 = 0.3; aa = 100/400 = 0. Allele frequencies p (^) A = 0.45 +1/2(0.3) = 0.6; q (^) a = 0.25 + 1/2(0.3) = 0. Expected H-W frequencies p 2 = (0.6) 2 = 0.36; 2 pq = 2(0.6)(0.4) = 0.0.48; q 2 = (0.4) 2 = 0. The HW expected genotype frequencies do not match the observed genotype frequencies so thispopulation is not in HW equilibrium. There is an excess of homozygotes and a deficit of heterozygotes. c. For how many generations will the effect of this single episode of admixture of populations bedetectable in a population established from population 3 with respect to genotype frequencies if mating is random with respect to this locus? (2 pts) After one generation of random mating this populations will be in HW proportions and will stay inthose proportions. You can only see the mixture in the first mixed generation.

  1. The frequency of phenylketonuria (PKU, caused by an autosomal recessive allele) is 0.00004 at birth. a. Assuming H-W, what is the frequency of the PKU allele? (1 pts) q^2 = 0.00004 so q = 0.006 and p = 1-0.006 = 0. b. What is the expected H-W ratio of PKU carriers (heterozygotes) to affected individuals (PKU homozygotes)? (1 pts) (^2) more than 300 carriers. pq / q^2 = 2(0.006)(0.994) / 0.00004 = 314; For every homozygous affected individual there will be